Problem: You have found the following ages (in years) of all 5 lizards at your local zoo: $ 2,\enspace 1,\enspace 1,\enspace 1,\enspace 3$ What is the average age of the lizards at your zoo? What is the standard deviation? You may round your answers to the nearest tenth.
Solution: Because we have data for all 5 lizards at the zoo, we are able to calculate the population mean $({\mu})$ and population standard deviation $({\sigma})$ To find the population mean , add up the values of all $5$ ages and divide by $5$ $ {\mu} = \dfrac{\sum\limits_{i=1}^{{N}} x_i}{{N}} = \dfrac{\sum\limits_{i=1}^{{5}} x_i}{{5}} $ $ {\mu} = \dfrac{2 + 1 + 1 + 1 + 3}{{5}} = {1.6\text{ years old}} $ Find the squared deviations from the mean for each lizard. Age $x_i$ Distance from the mean $(x_i - {\mu})$ $(x_i - {\mu})^2$ $2$ years $0.4$ years $0.16$ years $^2$ $1$ year $-0.6$ years $0.36$ years $^2$ $1$ year $-0.6$ years $0.36$ years $^2$ $1$ year $-0.6$ years $0.36$ years $^2$ $3$ years $1.4$ years $1.96$ years $^2$ Because we used the population mean $({\mu})$ to compute the squared deviations from the mean , we can find the variance $({\sigma^2})$ , without introducing any bias, by simply averaging the squared deviations from the mean $ {\sigma^2} = \dfrac{\sum\limits_{i=1}^{{N}} (x_i - {\mu})^2}{{N}} $ $ {\sigma^2} = \dfrac{{0.16} + {0.36} + {0.36} + {0.36} + {1.96}} {{5}} $ $ {\sigma^2} = \dfrac{{3.2}}{{5}} = {0.64\text{ years}^2} $ As you might guess from the notation, the population standard deviation $({\sigma})$ is found by taking the square root of the population variance $({\sigma^2})$ ${\sigma} = \sqrt{{\sigma^2}}$ $ {\sigma} = \sqrt{{0.64\text{ years}^2}} = {0.8\text{ years}} $ The average lizard at the zoo is 1.6 years old. There is a standard deviation of 0.8 years.